SUGGESTED SOLUTIONS FOR PROBLEM SET 5
SPRING 2011, MATH 312:01
Exercise 1. Suppose that f : R → R is periodic and integrable on every closed interval. If p is the
period of f , prove that for any a ∈ R,
Z a+p
Z p
f dx.
f dx =
a
0
0
Proof. Find n ∈ Z such that a = a − np and 0 ≤ a0 < p. Since f is p-periodic, a repeated
application of the change-of-variables theorem and the “algebra of integrable functions” theorem
yields:
Z p
Z a0
Z p
Z a0 +np
Z (n+1)p
f dx =
f dx +
f dx =
f dx +
f dx
0
a0
0
Z
=
a
Z
f dx +
np
Z p+a
=
a0 +np
np
(n+1)p
Z
p+a
f dx =
a
Z
f dx +
(n+1)p
(n+1)p
f dx
a
f dx,
a
as was to be shown.
Remark. In accordance with the section in which this exercise is placed, we note that the statement
can also be proved by using the Riemann sums with corresponding partitions and markings of the
two intervals, combined with the periodicity of f .
Exercise 2. Suppose g : [a, b] → R is bounded and continuous except at x1 , . . . , xn ∈ [a, b]. Prove
that g ∈ R(x) on [a, b].
Remark. Let us discuss the main ideas of the following (rather long) proof before indulging ourselves
in the nitty-gritty technical details. We have seen that the key to establish the Riemann-integrability
of a continuous function is that the function is uniformly continuous on closed intervals: that is,
given a small enough subinterval (say of length smaller than δ), the total variation of the function
is very small (within ε, for example).
When there are “a small enough number” of points of discontinuity, then some (but not all)
subintervals will contain those “bad points,” where the function is no longer uniformly continuous.
Therefore, we cannot contain the total variation of the function in a small range on those subintervals. Since the function is assumed to be bounded, however, the total variation is still within
some finite number. If we can bound the total length of the “bad” subintervals with a small enough
number, then the product of the “variation on the y axis” and the “variation on the x axis”—which
is what the “difference of the upper sum and the lower sum” is, essentially—will still be small.
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2
SPRING 2011, MATH 312:01
Proof. Fix ε > 0, and find an M > 0 such that |g(x)| < M for all x ∈ [a, b]. If ε ≥ 2M (b − a), then
finding a partition P such that U (P, f ) − L(P, f ) < ε is trivial. Therefore, we may well assume
that ε < 2M (b − a).
For each 1 ≤ i ≤ n we define an open interval Ii = (xi − ε/8M n, xi + ε/8M n), and let
!c
n
n
[
[
X = [a, b] r
Ii = [a, b] ∩
Ii .
i=1
i=1
[a, b] is closed, and each Ii is open, hence X is closed. Furthermore, X ⊆ [a, b], so that X is compact.
Note also that X is nonempty, for the sum of the length of each Ii (namely, (ε/4M n) · n = ε/4M )
is less than b − a.
g is continuous on the compact set X, hence g is uniformly continuous, and we may find δ > 0
such that |x − y| < δ implies |g(x) − g(y)| < ε/2(b − a) for all x, y ∈ X. Now, let P = {p0 , . . . , pm }
be any partition of [a, b] satisfying the following properties:
(1) The mesh of P is smaller than δ;
(2) P must contain both endpoints of each open interval Ii .
Take J to be the set of indices {1, . . . , n}. We shall divide J into two parts as follows: if the
subinterval [pj−1 , pj ] is entirely contained in X, then j ∈ Jg ; otherwise, j ∈ Jb .
We are now ready to make our estimate. We first observe that
U (P, f )−L(P, f ) =
m
X
X
X
(Mj −mj )(pj −pj−1 ) =
(Mj −mj )(pj −pj−1 )+
(Mk −mk )(pk −pk−1 ).
j=1
j∈Jg
k∈Jb
Since µ(P ) < δ, the “total variation” of f on each subinterval [pj−1 , pj ] ⊆ X is within ε/2(b − a).
More specifically, condition (1) stipulates that f is uniformly continuous on the intevral [pj−1 , pj ] of
length less than δ, hence any x, y ∈ [pj−1 , pj ] yields |g(x) − g(y)| < ε. Finally, [pj−1 , pj ] is compact,
whence we can find points cj , dj ∈ [pj−1 , pj ] such that g(cj ) = Mj and g(dj ) = mj . We may thus
conclude that
|Mj − mj | = |g(cj ) − g(dj )| < ε/2(b − a),
which immediately yields the inequality
X
X
(Mj − mj )(pj − pj−1 ) <
j∈Jg
j∈Jg
X
ε
ε
(pj − pj−1 ) =
pj − pj−1 .
2(b − a)
2(b − a)
j∈Jg
Jg ⊆ {1, . . . , n}, and each pj − pj−1 is postive, we have the inequality
X
j∈Jg
pj − pj−1 ≤
n
X
pj − pj−1 = b − a,
j=1
and it follows that
X
(Mj − mj )(pj − pj−1 ) <
j∈Jg
ε
ε
· (b − a) = .
2(b − a)
2
Now, condition (2) bounds the “total length” of the remaining subintervals. Indeed, (2) implies
that each subinterval furnished by P is either in X or in [a, b] r X. It then follows that the union
of the subintervals corresponding to the indices in Jb is precisely [a, b] r X, i.e.,
[
[a, b] r X =
[pj−1 , pj ].
j∈Jb
SUGGESTED SOLUTIONS FOR PROBLEM SET 5
3
Since [a, b]rX is the union of n intervals of length at most ε/4M n, we have the following inequality:
X
ε
ε
=
.
pj − pj−1 ≤ n ·
4M n
4M
j∈Jb
In addition, g is bounded by M on [a, b], so that |g(x) − g(y)| < 2M for any x, y ∈ [a, b]. Therefore,
X
X
ε
ε
(Mi − mi )(pj − pj−1 ) < 2M
pj − pj−1 ≤ 2M ·
= .
4M
2
j∈Jb
j∈Jb
It now follows that
U (P, f ) − L(P, f ) =
X
(Mj − mj )(pj − pj−1 ) +
j∈Jg
X
(Mk − mk )(pk − pk−1 ) <
k∈Jb
ε ε
+ = ε,
2 2
whence g ∈ R(x) on [a, b], as was to be shown.
Exercise 3. Suppose f and g are integrable on [a, b]. Define h(x) = max{f (x), g(x)}. prove that
h is integrable on [a, b].
Proof. Observe that
f (x) + g(x) |f (x) − g(x)|
+
2
2
for all x ∈ [a, b]. The desired result now follows from Theorem 5.12 and Theorem 5.9(i).
h(x) =
Remark. Analogously we have the following identity for any a, b ∈ R:
a + b |a − b|
−
.
2
2
Exercise 4. Assume f : [a, b] → R is continuous, f (x) ≥ 0 for all x ∈ [a, b], and
min{a, b} =
M = sup{f (x) : x ∈ [a, b]}.
Show that
"
#1/n ∞
Z b


[f (x)]n dx
a
n=1
converges to M .
Proof. If f is identically zero, then the equality holds trivially. We therefore assume that M > 0.
f is a continuous function on a compact interval [a, b], hence there exists m ∈ [a, b] such that
f (m) = M . We fix ε > 0, and pick a neighborhood (c, d) of m in [a, b] such that |f (x) − M | < ε for
all x ∈ (c, d). Observe that
Z d
Z d
Z b
Z b
(M − ε)n (d − c) =
(M − ε)n dx ≤
[f (x)]n dx ≤
[f (x)]n dx ≤
M n dx = M n (b − a).
c
c
a
a
The inequality we shall need is
b
Z
(M − ε)n (d − c) ≤
[f (x)]n dx ≤ M n (b − a).
a
Taking the nth root, we obtain
(M − ε)(d − c)1/n ≤
Z
a
b
!1/n
[f (x)]n dx
≤ M (b − a)1/n .
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SPRING 2011, MATH 312:01
As n → ∞, both (d − c)1/n and (b − a)1/n tend to 1, so that
!1/n
Z b
[f (x)]n dx
≤ M.
(M − ε) ≤ lim
n→∞
a
Since ε was arbitrary, we conclude that
Z
lim
n→∞
as was to be shown.
b
!1/n
[f (x)]n dx
= M,
a